2019 AMC 8 Problems/Problem 17

Problem 17

What is the value of the product $\left(\frac{1\cdot3}{2\cdot2}\right)\left(\frac{2\cdot4}{3\cdot3}\right)\left(\frac{3\cdot5}{4\cdot4}\right)\cdots\left(\frac{97\cdot99}{98\cdot98}\right)\left(\frac{98\cdot100}{99\cdot99}\right)?$

$\textbf{(A) }\frac{1}{2}\qquad\textbf{(B) }\frac{50}{99}\qquad\textbf{(C) }\frac{9800}{9801}\qquad\textbf{(D) }\frac{100}{99}\qquad\textbf{(E) }50$

Solution 1(Telescoping)

We rewrite: $\frac{1}{2}\cdot\left(\frac{3\cdot2}{2\cdot3}\right)\left(\frac{4\cdot3}{3\cdot4}\right)\cdots\left(\frac{99\cdot98}{98\cdot99}\right)\cdot\frac{100}{99}$

The middle terms cancel, leaving us with

$\left(\frac{1\cdot100}{2\cdot99}\right)= \boxed{\textbf{(B)}\frac{50}{99}}$

~ hpotter0104

Solution 2

If you calculate the first few values of the equation, all of the values tend to $\frac{1}{2}$ , but are not equal to it. The answer closest to $\frac{1}{2}$ but not equal to it is $\boxed{\textbf{(B)}\frac{50}{99}}$ .~hpotter0104

Solution 3

Rewriting the numerator and the denominator, we get $\frac{\frac{100! \cdot 98!}{2}}{\left(99!\right)^2}$ . We can simplify by canceling 99! on both sides, leaving us with: $\frac{100 \cdot 98!}{2 \cdot 99!}$ We rewrite $99!$ as $99 \cdot 98!$ and cancel $98!$ , which gets $\boxed{\frac{50}{99}}$ . Answer B.

Video Solution

https://www.youtube.com/watch?v=ffHl1dAjs7g&list=PLLCzevlMcsWNBsdpItBT4r7Pa8cZb6Viu&index=1 ~ MathEx

Video Solution - https://youtu.be/Lw8fSbX_8FU (Also explains problems 11-20)

2019 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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